What is the extraneous solution to these equations? $\dfrac{x^2 - 5}{x - 6} = \dfrac{7x - 11}{x - 6}$
Answer: Multiply both sides by $x - 6$ $ \dfrac{x^2 - 5}{x - 6} (x - 6) = \dfrac{7x - 11}{x - 6} (x - 6)$ $ x^2 - 5 = 7x - 11$ Subtract $7x - 11$ from both sides: $ x^2 - 5 - (7x - 11) = 7x - 11 - (7x - 11)$ $ x^2 - 5 - 7x + 11 = 0$ $ x^2 + 6 - 7x = 0$ Factor the expression: $ (x - 6)(x - 1) = 0$ Therefore $x = 6$ or $x = 1$ At $x = 6$ , the denominator of the original expression is 0. Since the expression is undefined at $x = 6$, it is an extraneous solution.